Name ID Number TA Section Time MTH 234 Exam 2 Practice October 19 2010 50 minutes Sects 13 1 13 3 14 1 14 7 1 No calculators or any other devices allowed If any question is not clear ask for clarification No credit will be given for illegible solutions If you present different answers for the same problem the worst answer will be graded Show all your work Box your answers a 15 points Find the position r and velocity vector functions v of a particle that moves with an acceleration function a t h0 0 10i m sec2 knowing that the initial velocity and position are given by respectively v 0 h0 1 2i m sec and r 0 h0 0 3i m b 5 points Draw an approximate picture of the graph of r t for t 0 Solution a a t h0 0 10i v t hv0x v0y 10 t v0z i v 0 h0 1 2i v0x 0 v0y 1 v0z 2 v t h0 1 10 t 2i r t hr0x t r0y 5t2 2t r0z i r 0 h0 0 3i r t h0 t 5t2 2t 3i b z 3 parabola y r0x 0 r0y 0 r0z 3 2 a 10 points Find and sketch the domain of the function f x t ln 3x 2t b 10 points Find all possible constants c such that the function f x t above is solution of the wave equation ftt c2 fxx 0 Solution a The argument in the ln function must be positive then the domain is t t 3 2 x D x t R 2 3x 2t 0 2 x 3 b 2 3x 2t 4 ftt 3x 2t 2 ft 0 ftt cfxx 3 3x 2t 9 3x 2t 2 fx fxx 4 9 1 c2 4 9c2 2 2 2 3x 2t 3x 2t 3x 2t 9c2 4 c 2 3 3 a 10 points Find the direction in which f x y increases the most rapidly and the directions in which f x y decreases the most rapidly at P0 and also find the value of the directional derivative of f x y at P0 along these directions where f x y x3 e 2y and P0 1 0 b 10 points Find the directional derivative of f x y above at the point P0 in the direction given by v h1 1i Solution a The direction in which f increases the most rapidly is given by f and the one in which decreases the most rapidly is f So f x y h3x2 e 2y 2x3 e 2y i f 1 0 h3 2i f 1 0 h 3 2i The value of the directional derivative along these directions is respectively f 1 0 and f 1 0 where f 1 0 9 4 13 1 b A unit vector along h1 1i is u h1 1i then 2 1 5 Du f 1 0 f 1 0 u h3 2i h1 1i 2 2 5 Du f 1 0 2 4 a 10 points Find the tangent plane approximation of f x y x cos y 2 y 2 e x at the point 0 1 b 10 points Use the linear approximation computed above to approximate the value of f 0 1 0 9 Solution a f x y x cos y 2 y 2 e x fx x y cos y 2 y 2 e x fy x y x sin y 2 2ye x 2 f 0 1 1 fx 0 1 cos 2 1 1 fy 0 1 2 Then the linear approximation L x y is given by L x y x 0 2 y 1 1 L x y x 2y 1 b The linear approximation of f 0 1 0 9 is L 0 1 0 9 which is given by L 0 1 0 9 0 1 2 0 1 1 0 1 1 1 1 L 0 1 0 9 1 1 5 20 points Find every local and absolute extrema of f x y x2 3y 2 2y on the unit disk x2 y 2 1 and indicate which ones are the absolute extrema In the case of the interior stationary points decide whether they are local maximum minimum of saddle points Solution We first compute the interior stationary points which are x y solutions of f h2x 6y 2i h0 0i 1 y 3 x 0 The point 0 1 3 belongs to the disk x2 y 2 1 so we have to decide whether it is a local maximum minimum or saddle point fxx 2 fyy 6 fxy 0 2 D fxx fyy fxy 12 0 fxx 0 1 0 3 is a local minimum This point is also a candidate for absolute minimum so we record the value of f 1 1 1 3 2 0 f 0 0 3 3 9 3 3 We now look for extreme point on the boundary x2 y 2 1 Wepevaluate f x y along the boundary From the equation x2 y 2 1 we compute x 1 y 2 This function is differentiable for y 1 1 but is not differentiable at y 1 Since we need to use the chain rule to find the extrema of g y f x y y and the chain rule does not hold at y 1 we need to consider these points 0 1 separately 0 1 f 0 1 5 0 1 f 0 1 1 Now we find local extrema on g y f x y y in the interval y 1 1 The function g is given by g y 1 y 2 3y 2 2y g y 1 2y 2 2y 0 The local extrema for g are the points y solutions p of g y 0 that we conclude y 1 2 and x 1 1 4 3 2 that is 3 1 3 1 3 3 1 f 2 2 2 2 2 4 4 2 is 4y 2 0 so 1 2 Therefore the absolute extrema are 0 1 absolute maximum 1 0 3 absolute minimum